Robust Control ÉCOLE POLYTECHNIQUE FÉDÉRALE DE LAUSANNE. Alireza Karimi Laboratoire d Automatique
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1 Robust Control ÉCOLE POLYTECHNIQUE FÉDÉRALE DE LAUSANNE Alireza Karimi Laboratoire d Automatique
2 Course Program 1. Introduction 2. Norms for Signals and Systems 3. Basic Concepts (stability and performance) 4. Uncertainty and Robustness 5. Stabilization (coprime factorization) 6. Design Constraints 7. Loopshaping 8. Model Matching 10. Linear Fractional Transformation 11. H Control 12. Model and Controller Reduction 13. Robust Control by Convex Optimization 14. LMIs in Robust Control 15. Robust Pole Placement 16. Parametric uncertainty 9. Design for Performance References: Feedback Control Theory by Doyle, Francis and Tannenbaum (on the website of the course) Essentials of Robust Control by Kemin Zhou with Doyle, Prentice-Hall, 1998 Robust Control 2
3 Introduction Classical Control Robust Control u(t) y(t) P0 u(t) P0 + y(t) P = P 0 + Uncertainty sources: P 0 : nominal model : plant uncertainty Structured: parametric uncertainty, multimodel uncertainty Unstructured: frequency-domain uncertainty, unmodeled dynamics, nonlinearity Robust Control Objective: Design a controller satisfying stability and performance for a set of models Robust Control 3
4 Model Uncertainty and Feedback Feedback control has been used for the first time to overcome the model uncertainty r(t) e(t) u(t) C P + - v(t) y(t) T = CP 1+CP S = 1 1+CP Transfer function between r and y Transfer function between v and y For very large CP, T 1 (tracking) and S 0 (disturbance rejection) whatever the plant model is. For an open-loop stable system: C =0robust stability C robust performance Loopshaping: C(jω)P (jω) should be large in the frequencies where good performances are desired and small where the stability is critical Introduction 4
5 Norms for Signals and Systems Norms for signals: Consider piecewise continuous signals mapping (, + ) to R. A norm must have the following four properties: 1. u 0 (positivity) 2. au = a u, a R (homogenity) 3. u =0 u(t) =0 t (positive definiteness) 4. u + v u + v (triangle inequality) 1-Norm: u 1 = u(t) dt 2-Norm: u 2 = Robust Control ( u 2 (t)dt) 1/2 u 2 2 -Norm: u =sup u(t) t ( 1/p p-norm: u p = u(t) dt) p 1 p is the total signal energy 5
6 Norms for Signals Average power of a signal is denoted by: pow(u) = ( lim T 1 2T T T u 2 (t)dt) 1/2 pow is not a norm (it is not positive definite) Remark: One says u(t) L p if u p < where L p is an infinite-dimensional Banach space (L 2 is an infinite-dimensional Hilbert space as well). Recall: A Banach space is a complete vector space with a norm. A Hilbert space is a complete vector space with an inner product <x,y>such that the norm defined by x = <x,x>. A Hilbert space is always a Banach space, but the converse need not hold. Examples: 1(t) L but / L 1,/ L 2 u(t) =(1 e t )1(t) L but / L 1,/ L 2 u(t) =e t 1(t) L, L 1, L 2 (Besides, u(t) is a power signal pow(u) =0) u(t) =sin(t) L but / L 1,/ L 2 (u(t) is a power signal) Norms for Signals and Systems 6
7 Norms for Systems We consider linear, time-invariant, causal and usually finite-dimensional systems. y(t) =G(t) u(t), y(t) = Some definitions: G(t τ)u(τ)dτ, Ĝ(s) is stable if it is analytic in the closed RHP (Re s 0) Ĝ(s) is proper if Ĝ(j ) is finite (deg den deg num) Ĝ(s) is strictly proper if Ĝ(j ) =0deg den > deg num Ĝ(s) is biproper if (deg den = deg num) Norms for SISO systems: Ĝ(s) =L[G] 2-Norm: Ĝ 2 = ( 1 2π Parsval s theorem: (for stable systems) Ĝ 2 = ( 1 2π Norms for Signals and Systems Ĝ(jω) 2 dω) 1/2 -Norm: Ĝ =sup ω Ĝ(jω) 2 dω) 1/2 = ( ) 1/2 G(t) 2 dt Ĝ(jω) 7
8 Norms for Systems Remarks: L 2 is a Hilbert space of scalar-valued functions on jr. The inner product for this Hilbert space is defined as: < ˆF,Ĝ>= 1 2π ˆF (jω)ĝ(jω)dω We say Ĝ(s) L 2 if Ĝ 2 <. It is the case iff Ĝ is strictly proper and has no poles on the imaginary axis. We say Ĝ(s) L if Ĝ <. It is the case iff Ĝ is proper and has no poles on the imaginary axis. L is a Banach space of scalar-valued functions on jr. Ĝ is the peak value of the Bode magnitude plot of Ĝ. It is also the distance from the origin to the farthest point on the Nyquist plot of Ĝ. H 2 and H are respectively subspaces of L 2 and L with Ĝ(s) stable (H p spaces are usually called Hardy spaces). Examples: 1 s 1 L 2, L but / H 2, H s+1 s+2 H but / H 2 1 s 2 +1 / L 2, L Norms for Signals and Systems 8
9 Norms for Systems Norms for matrices: 1-Norm: The maximum absolute column sum norm is defined as A 1 =max j n i=1 a ij. 2-Norm: The spectral norm or simply the norm of A is defined as: A 2 = λ max (A A). -Norm: The maximum absolute row sum norm is defined as A =max i n j=1 a ij. F-Norm: Frobenius norm is defined as A F = trace(a A) Induced p-norm: The induced p-norm is defined from a vector p-norm: A p =max x 0 Ax p x p Remarks: The induced 2-norm and the norm of A are the same and called also the natural norm. This norm is also equal to the maximum singular value of A. A = σ(a). The spectral radius ρ(a) = λ max (A) is not a norm. Norms for Signals and Systems 9
10 Norms for Systems Norms for MIMO systems: Given Ĝ(s) a multi-input multi-output system 2-Norm: This norm is defined as Ĝ 2 = ( 1 2π trace [Ĝ (jω)ĝ(jω) ] ) 1/2 dω -Norm: The H norm is defined as Ĝ =sup ω Ĝ(jω) =sup σ[ĝ(jω)] ω Remark: The infinity norm has an important property (submultiplicative) ĜĤ Ĝ Ĥ Norms for Signals and Systems 10
11 Computing the Norms How to compute the 2-norm: Suppose that Ĝ L 2,wehave: Ĝ 2 2 = 1 2π Ĝ(jω) 2 dω = 1 2πj j j Ĝ( s)ĝ(s)ds = 1 2πj Ĝ( s)ĝ(s)ds Then by the residue theorem, Ĝ 2 2 left half-plane (LHP). equals the sum of the residues of Ĝ( s)ĝ(s) at its poles in the How to compute the -norm: Choose a fine grid of frequency point {ω 1,...,ω N }, then an estimate for Ĝ is: max Ĝ(jω k ) 1 k N or for the MIMO systems: max σ[ĝ(jω k)] 1 k N Matlab commands: norm, bode, frsp, vsvd Norms for Signals and Systems 11
12 Computing the Norms State-space methods for 2-norm: Consider a SISO state-space model H 2 ẋ(t) =Ax(t)+Bu(t) y(t) =Cx(t) L = sˆx(s) =Aˆx(s)+Bû(s) ŷ(s) =C ˆx(s) Ĝ(s) =C(sI A) 1 B = impulse response : G(t) =Ce ta B From Parseval s theorem: Ĝ 2 2 = G 2 2 = 0 ( Ce ta B ) ( B T e tat C T ) dt = CLC T where L = 0 e ta BB T e tat dt is the observability Gramian and can be obtained from following Lyapunov equation: AL + LA T + BB T =0and the 2-norm is Ĝ 2 =(CLC T ) 1/2 For MIMO systems we have Ĝ 2 = [ trace(clc T ) ] 1/2 Norms for Signals and Systems 12
13 Computing the Norms State-space methods for -norm: Consider a SISO strictly proper state-space model L. Theorem: Ĝ <γiff the Hamiltonian matrix H has no eigenvalues on the imaginary axis: H = A γ 2 BB T C T C A T Bisection algorithm: 1. Select γ u and γ l such that γ l Ĝ γ u ; 2. If (γ u γ l )/γ l specified level., stop and Ĝ (γ u + γ l )/2. Otherwise continue; 3. Set γ =(γ u + γ l )/2 and test if Ĝ <γ 4. If H has no eigenvalue on jr, set γ u = γ otherwise set γ l = γ; go back to step 2. For MIMO biproper (D 0) systems: H = A + BR 1 D T C BR 1 B T and R = γ 2 I D T D C T (I + DR 1 D T )C (A + BR 1 D T C) T Norms for Signals and Systems 13
14 Input-output relationships If we know how big the input is, how big is the output going to be? Output Norms for Two Inputs u(t) δ(t) sin(ωt) y 2 Ĝ 2 y G Ĝ(jω) Proofs: If u(t) =δ(t) then y(t) = G(t τ)δ(τ)dτ = G(t), so y 2 = G 2 = Ĝ 2 If u(t) =δ(t) then y(t) =G(t), so y = G If u(t) =sin(ωt) then y(t) = Ĝ(jω) sin(ωt + φ), so y 2 = and y = Ĝ(jω) Norms for Signals and Systems 14
15 Input-output relationships Proofs: System Gain: sup{ y : u 1} u(t) L 2 u(t) L y 2 Ĝ y Ĝ 2 G 1 Entry (1,1): We have y 2 2 = ŷ 2 2 = 1 2π Ĝ(jω) 2 û(jω) 2 dω Ĝ 2 1 2π û(jω) 2 dω = Ĝ 2 û 2 2 = Ĝ 2 u 2 2 Entry(2,1): According to the Cauchy-Schwartz inequality y(t) = G(t τ)u(τ)dτ ( ) 1/2 ( G 2 (t τ)dτ ) 1/2 u 2 (τ)dτ = G 2 u 2 = Ĝ 2 u 2 y Ĝ 2 u 2 Norms for Signals and Systems 15
16 Basic Concepts Basic Feedback Loop: r - x 1 C v u F d x 2 P x 3 y n x 1 = r Fx F x 1 r x 2 = d + Cx 1 = C 1 0 x 2 = d x 3 = n + Px 2 0 P 1 x 3 n 1 x F r 1 PF F r x 2 = C 1 0 d = 1 1+PCF C 1 CF d 0 P 1 n PC P 1 n x 3 Well-posedness: The system is well-posed iff the above matrix is nonsingular. A stronger notion of well-posedness is that all the nine transfer functions be proper. A necessary and sufficient condition for this is that 1+PCF not be strictly proper (PCF( ) 1) Robust Control 16
17 Internal Stability If the following nine transfer functions are stable then the feedback system is internally stable. 1 1+PCF 1 PF F C 1 CF PC P 1 Theorem: The feedback system is internally stable iff P = N P M P, there are no zeros in Re s 0 in the characteristic polynomial or the following two conditions hold: N P N C N F + M P M C M F =0 (a) The transfer function 1+PCF has no zeros in Re s 0. C = N C M C, F = N F M F (b) There is no pole-zero cancellation in Re s 0 when the product P CF is formed. Basic Concepts 17
18 Asymptotic Tracking Internal Model Principle: For perfect asymptotic tracking of r(t), the loop transfer function ˆL = ˆP Ĉ (with ˆF =1) must contain the unstable poles of ˆr(s). Theorem: Assume that the feedback system is internally stable and n=d=0. (a) If r(t) is a step, then lim t e(t) =r(t) y(t) =0iff Ŝ =(1+ˆL) 1 has at least one zero at the origin. (b) If r(t) is a ramp, then lim t e(t) =0iff Ŝ has at least two zeros at the origin. (c) If r(t) =sin(ωt), then lim t e(t) =0iff Ŝ has at least one zero at s = jω. Final-Value Theorem: If ŷ(s) has no poles in Re s 0 except possibly one pole at s = 0 then: lim y(t) = lim sŷ(s) t s 0 Proof (a): We have ˆr(s) = c s and ê(s) =Ŝ(s) c then lim e(t) = lim s sŝ(s) c t s 0 s iff Ŝ has at least one zero at origin. For this, ˆP or Ĉ should have a pole at origin.. The limit is zero Basic Concepts 18
19 Performance Tracking performance can be quantified in terms of a weighted norm of the sensitivity function 1 Sensitivity Function: Transfer function from r to tracking error e : S = 1+PC Complementary Sensitivity Function: Transfer function from r to y : T = PC 1+PC S is the relative sensitivity of T with respect to relative perturbations in P: S = lim P 0 T/T P/P = dt dp P T C(1 + PC) PC2 P (1 + PC) = (1 + PC) 2 PC = 1 1+PC Performance Specification: 1. r(t) is any sinusoid of amplitude 1 filtered by W 1, then the max. amp. of e is W 1 S. 2. Suppose that {r = W 1 r pf, r pf 2 1}, then sup r e 2 = W 1 S. 3. In some applications good performance is achieved if S(jω) < W 1 (jω) 1, ω or W 1 S < 1 W 1 (jω) < 1+L(jω), ω Basic Concepts 19
20 Uncertainty and Robustness Plant Uncertainty: We cannot exactly model the physical systems so there is always the modeling errors. The best technic is to define a model set which can be structured or unstructured. Structured model set such as parametric uncertainty or multiple model set P = { 1 s 2 + as +1 : a min a a max } or P = {P 0,P 1,P 2,P 3 } Unstructured model set such as unmodeled dynamics or disk uncertainty P = {P 0 + : γ} Conservatism: Controller design for a model set greater than the real model set leads to a conservative design. Uncertainty Models (unstructured): Additive uncertainty Multiplicative uncertainty Feedback uncertainty P P P = P + W 2 P = P (1 + W2 ) P = 1+ W 2 P or P = 1+ W 2 P : true model P : nominal model : norm-bounded uncertainty W 2 : weighting filter Robust Control 20
21 Examples Example 1: k frequency-response models are identified. Find the multiplicative uncertainty model and the weighting filter. if P = P (1 + W 2 ) P P 1= W 2 P (jω) 1 P (jω) 1 W 2(jω) Let (M ik,φ ik ) be the magnitude-phase at ω i in k-th experiment and (M i,φ i ) that of the nominal model (e.g. the mean value). max k M ik e φ ik M i e φ i Example 2: Suppose that P (s) ={ k s 2 multiplicative uncertainty. P (s) = k 0 s 2 1 W 2(jω i ) i :0.1 k 10}. Represent this model by the P (jω) P (jω) 1 W 2(jω) max 0.1 k 10 k 1 k 0 W 2(jω) The best value for k 0 is 5.05 which gives W 2 (s) =4.95/5.05 Uncertainty and Robustness 21
22 Examples Example 3: Assume that P (s) = 1 s 2 and P (s) =e τs 1 s 2 where 0 τ 0.1. Find W 2 (s) for the multiplicative uncertainty model. P (jω) P (jω) 1 W 2(jω) e τjω 1 W 2 (jω) Using the Bode diagram we can find W 2 (s) = 0.21s 0.1s+1. ω, τ Example 4: Consider the model set { 1 s 2 +as+1 :0.4 a 0.8}. Find W 2(s) for Feedback uncertainty model. Take: a = , 1 1 So where P (s) = Uncertainty and Robustness 1 s s +0.2 s +1 = P (s) = 1 s s +1, P (s) 1+ W 2 (s)p (s) W 2(s) =0.2s 22
23 Robust Stability Robustness: A controller is robust with respect to a closed-loop characteristic, if this characteristic holds for every plant in P Robust Stability: A controller is robust in stability if it provides internal stability for every plant in P Stability margin: For a given model set with an associate size, it can be defined as the largest model set stabilized by a controller. Stability margin for an uncertainty model: Given P = P (1 + W 2 ) with β, the stability margin for a controller C is the least upper bound of β. Im L Modulus margin: The distance from -1 to the open-loop Nyquist curve. M m = inf = ω [ sup ω 1 L(jω) =inf 1 1+L(jω) ω ] 1 = S 1 1+L(jω) 1 M m L(jω) 1 Re L Uncertainty and Robustness 23
24 Robust Stability Small Gain Theorem: Suppose H RH (s) and let γ>0. The following feedback loop is internally stable for all (s) RH with 1/γ if and only if H <γ H(s) Remark: For a given with 1/γ the condition H <γis only sufficient and very conservative. However for all RH, it is a necessary and sufficient condition. Robust stability condition for plants with additive uncertainty: P = P + W 2 H = W 2 C 1+CP Closed-loop system is internally stable for all 1 iff W 2 CS < 1 r(t) W 2 + C P - y(t) Uncertainty and Robustness 24
25 Robust Stability Robust stability condition for plants with multiplicative uncertainty: P = P (1+ W 2 ) H = W 2 CP 1+CP Closed-loop system is internally stable for all 1 iff W 2 T < 1 r(t) W 2 + C P - y(t) Proof: Assume that W 2 T < 1. We show that the winding number of 1+CP around zero is equal to that of 1+C P. 1+C P =1+CP(1 + W 2 )=1+CP + CP W 2 =1+CP +(1+CP)T W 2 1+C P =(1+CP)(1 + W 2 T ) so Wno { (1 + C P )} = Wno{(1 + CP)} + Wno{(1 + W 2 T )}. But Wno {(1 + W 2 T )} =0because W 2 T < 1 Uncertainty and Robustness 25
26 Robust Stability Robust stability condition for plants with feedback uncertainty (1): P = P 1 H = W 2 1+ W 2 1+CP Closed-loop system is internally stable for all 1 iff W 2 S < 1 r(t) W 2 - C P y(t) - Robust stability condition for plants with feedback uncertainty (2): P = P 1+ W 2 P H = W P 2 1+CP Closed-loop system is internally stable for all 1 iff W 2 PS < 1 r(t) - C W 2 - P y(t) Uncertainty and Robustness 26
27 Robust Performance Nominal performance condition: W 1 S < 1 Robust stability condition for multiplicative uncertainty: W 2 T < 1 Robust performance for multiplicative uncertainty: W 2 T < 1 and W 1 S < 1 where: 1 S = 1+C P = 1 1+CP(1 + W 2 ) = 1 (1 + CP)(1 + W 2 T ) = S 1+ W 2 T Robust performance conditions: W 2 T < 1 and W 1 S 1+ W 2 T < 1 Theorem: A necessary and sufficient condition for robust performance is W 1 S + W 2 T < 1 Robust performance for additive uncertainty: W 2 CS < 1 and W 1 S < 1 where: 1 S = 1+C P = 1 S = 1+CP + C W 2 1+ W 2 CS W 1 S 1+ W 2 CS < 1 Or equivalently in one inequality condition: W 1 S + W 2 CS < 1 Uncertainty and Robustness 27
28 Stabilization The main objective is to parameterize all of the controllers which provide internal stability for a given plant Theorem: Assume that P RH (P is stable). The set of all stabilizing controllers is given by: { } Q C := 1 PQ Q RH Proof: (F =1) 1 PF F 1 1+PCF C 1 CF = PC P 1 1 PQ P (1 PQ) 1(1 PQ) Q 1 PQ Q RH PQ P(1 PQ) 1 PQ On the other hand, suppose that C stabilizes P then define Q := C 1+CP RH which leads to C = Q 1 PQ In this parameterization sensitivity and complementary sensitivity are Robust Control S =1 PQ T = PQ 28
29 Coprime Factorization Objective: Given P, find M,N,X and Y RH such that: P = N M NX + MY =1 Remarks: N and M are called coprime factors of G over RH N and M can have no common zeros in Re s 0 nor at s = N(s 0 )X(s 0 )+M(s 0 )Y (s 0 )=0 1 If P is stable we have : M =1,N = P, X =0,Y =1 It is easy to obtain N and M, for example: P (s) = 1 s 1 = N(s) M(s) N(s) = if k>1 then M and N have a common zero at s =, sok =1 1 (s +1) k, M(s) = s 1 (s +1) k Stabilization How to compute X(s) and Y (s)? 29
30 Coprime Factorization Euclid s algorithm: Given polynomials m(λ) and n(λ) (deg n deg m ) find polynomials x(λ) and y(λ) such that nx + my =1. Step 1: Divide m into n to get quotient q 1 and remainder r 1 : n = mq 1 + r 1, deg r 1 < deg m Step 2: Divide r 1 into m to get quotient q 2 and remainder r 2 : m = r 1 q 2 + r 2, deg r 2 < deg r 1 Step 3: Divide r 2 into r 1 to get quotient q 3 and remainder r 3 : r 1 = r 2 q 3 + r 3, deg r 3 < deg r 2 Continue Stop at step k when r k is a nonzero constant. Find r 3 as a function of m, n and q i : r 2 {}}{ r 3 =(n mq 1 ) (m (n mq 1 ) q 2 ) q 3 = n(1 + q 2 q 3 )+m( q 3 q 1 q 1 q 2 q 3 ) }{{}}{{} r 1 r 1 which gives: x = 1 r 3 (1 + q 2 q 3 ) and y = 1 r 3 ( q 3 q 1 q 1 q 2 q 3 ) Stabilization 30
31 Coprime Factorization Procedure to find M,N,X and Y for an unstable plant G: Step 1: Transform G(s) to G(λ) under the mapping s =(1 λ)/λ. Write G = n(λ) m(λ) Step 2: Using Euclid s algorithm, find x(λ) and y(λ) such that: nx + my =1 Step 3: Find M,N,X and Y from m, n, x and y under the mapping λ =1/(s +1) State-Space Method: Step 1: Transform G(s) to A, B, C and D (state space realization) Step 2: Compute F and H so that A + BF and A + HC are stable (F=-place(A,B,Pf)) Step 3: Compute M,N,X and Y as follows: M(s) := A + BF B N(s) := F 1 X(s) := A + HC H Y (s) := F 0 A + BF C + DF A + HC B D B HD F 1 Stabilization 31
32 Controller Parametrization Theorem: The set of all Cs for which the feedback system is internally stable equal: C = { X + MQ Y NQ : Q RH Proof: For C = N c M c, the stability condition is: (NN c + MM c ) 1 RH, but we have: N(X + MQ)+M(Y NQ)=NX + MY =1 (NN c + MM c ) 1 RH } Conversely, if C stabilizes the closed-loop system we should show that it belongs to the above set. C is stabilizing V := (NN c + MM c ) 1 RH NN c V + MM c V =1 Let Q be the solution of M c V = Y NQ. From the above equation and NX + MY =1we find that N c V = X + MQ so the controller C = N cv the set of all stabilizing controller. It is easy M c V to verify that Q RH Remark: The sensitivity functions are: Stabilization S = 1 1+CP = M(Y NQ) T = CP 1+CP = N(X + MQ) 32
33 Example Let Compute a proper controller C so that: P (s) = 1 (s 1)(s 2) 1. The feedback system is internally stable. 2. Perfect asymptotic tracking of step reference (d =0). 3. Perfect asymptotic disturbance rejection when d =sin10t (r =0). Procedure: Parameterize all stabilizing controllers. Reduce the asymptotic specs to interpolation constraints on the parameters. Find (if possible) a parameter to satisfy these constraints. Back-substitute to get the controller. Stabilization 33
34 Design Constraints Algebraic Constraints: S + T =1so S(jω) and T (jω) cannot both be less than 1/2 at the same frequency. A necessary condition for robust performance is that: min{ W 1 (jω), W 2 (jω) } < 1, ω So at every frequency either W 1 or W 2 must be less than 1. Typically W 1 is monotonically decreasing and W 2 is monotonically increasing. If p is a pole and z a zero of L both in Re s 0 then: S(p) =0 S(z) =1 T (p) =1 T (z) =0 Analytic Constraints: Bounds on the weights W 1 and W 2 : W 1 S W 1 (z) W 2 T W 2 (p) Proof from the Maximum Modulus Theorem: F = Robust Control sup F (s) Re s>0 34
35 Analytic Constraints All-Pass and Minimum-Phase Transfer Functions: F (s) RH is all-pass if F (jω) =1 ω G(s) RH is minimum-phase if it has no zeros in Re s>0. It has the minimum phase among all transfer functions with the same magnitude (FG where F is all-pass). Every function G in RH can be presented as G = G ap G mp Suppose that L = CP has no poles on the imaginary axis, so S =(1+L) 1 = S ap S mp and S mp has no zeros on the imaginary axis. Thus S 1 mp RH. Suppose that z and p are the only zero and pole of P in the closed RHP and C has neither poles nor zeros there. Then: Design Constraints S ap = s p s + p S(z) =1 S mp (z) =Sap 1 (z) = z + p z p W 1(z) z + p z p Then: W 1 S = W 1 S mp W 1 (z)s mp (z) = Similarly: T ap = s z s + z and T (p) =1 W 2T W 2(p) p + z p z 35
36 Analytic Constraints Example: Consider the inverse pendulum problem. Linearized model: x = θ (M + m)ẍ + ml( θ cos θ θ 2 sin θ) = u m(ẍ cos θ + l θ g sin θ) = d 1 s 2 [Mls 2 (M + m)g] ls2 g ls 2 s 2 M+m m s2 u d y x u d M θ l m T ux = ls 2 g s 2 [Mls 2 (M + m)g] T uθ = 1 Mls 2 (M + m)g RHP poles and zeros: z = g/l p =0, 0, T uy = g s 2 [Mls 2 (M + m)g] (M + m)g no RHP zero For T ux if m M W 2 T 1 ( W 2 (p) is an increasing function) the system is difficult to control. The best case is m/m and l large. Ml For T uθ and T uy a larger l gives a smaller p so the system is easier to stabilize. Design Constraints 36
37 Analytic Constraints The Waterbed Effect Lemma: For every point s 0 = σ 0 + jω 0 with σ 0 > 0, log S mp (s 0 ) = 1 π σ 0 log S(jω) σ0 2 +(ω ω 0) 2 dω Theorem: Suppose that P has a zero at z with Re z>0 and: M 1 := max S(jω) ω 1 ω ω 2 M 2 := S Then there exist positive constants c 1 and c 2, depending only on ω 1,ω 2 and z, such that : c 1 log M 1 + c 2 log M 2 log S 1 ap (z) 0 Theorem (The Area Formula): Assume that the relative degree of L is at least 2. Then 0 log S(jω) dω = π(log e) i Re p i where {p i } denotes the set of poles of L in Re s>0. Design Constraints 37
38 Loopshaping Objective: Given P, W 1 and W 2 find controller C providing internal stability and robust performance: W 1 S + W 2 T < 1 or Γ(jω):= W 1 (jω) 1+L(jω) + W 2 (jω)l(jω) 1+L(jω) < 1 ω Idea: Find graphically L(jω) satisfying the above condition and then compute C = L/P Note that we assume P is minimum phase and stable. We have: Γ 1+L = W 1 + W 2 L and 1 L 1+L 1+ L W 1 + W 2 L 1+ L Γ W 1 + W 2 L 1 L So if W 1 + W 2 L < 1 L Γ < 1: In low frequencies L > 1 L > W W 2 W 1 1 W 2 W 1 1 > W 2 In high frequencies L < 1 L < 1 W 1 1+ W 2 1 W 1 W 2 Robust Control W 2 1 > W 1 38
39 Procedure step 1: Plot two curves on log-log scale: at LF ( W 1 > 1 > W 2 ) W 1 1 W 2 and at HF ( W 2 > 1 > W 1 ) 1 W 1 W 2 step 2: Fit the graph of L on the same plot such that: at low frequency it lies above the first curve and also 1 at high frequency it lies below the second curve and 1 at very high frequency let it roll off at least as fast as does P (so C is proper) near crossover frequency do a smooth transition, keeping the slope as gentle as possible. Because the slope of L determines the phase of L (Bode s integral): L(jω 0 )= 1 π d ln L dν ln coth ν 2 dν where ν =ln(ω/ω 0) The steeper the graph of L near the crossover frequency, the smaller the value of L and larger the phase margin step 3: Get a stable, minimum-phase TF for L such that L(0) > 0 and compute C = L/P Loopshaping 39
40 Example Assume that the relative degree of P equals 1. Find L for robust performance if the objective is to track sinusoidal signals over the frequency range from 0 to 1 rad/s and the weighting function W 2 is: Loopshaping W 2 (s) = s +1 20(0.01s +1) We can define W 1 as follows (in loopshaping design it is not necessary to have a rational TF for W 1 ): a 0 ω 1 W 1 (jω) = The larger the value of a, the smaller the tracking error 0 else LF ( W 1 > 1): ω<1 HF ( W 2 > 1): ω 20 W 1 Plot 1 W 2 in LF (ω 1 W 1 <1) and in HF (ω >20) W 2 Choose L = b s +1 and find b such that in HF L 1 W 1 W 2 = 1 W 2 ( b 20) Find the maximum value of a such that in LF L W 1 1 W 2 = a 1 W 2 a =
41 Model Matching Objective: Given T 1 (s) and T 2 (s), stable proper transfer functions, find a stable Q(s) to Q T 2 minimize T 1 T 2 Q Trivial case: If T 1 /T 2 is stable then the unique r(t) T 1 - ε(t) optimal Q is T 1 /T 2 and γ opt =min T 1 T 2 Q =0 Simplest nontrivial case: T 2 has only one RHP zero at s = s 0. Then by the maximum modulus theorem: T 1 T 2 Q T 1 (s 0 ) T 2 (s 0 )Q(s 0 ) = T 1 (s 0 ) γ opt T 1 (s 0 ) Note that Q = T 1 T 1 (s 0 ) T 2 is stable and leads to γ opt = T 1 (s 0 ). Example: T 1 (s) = 4 s+3, T 2(s) = s 2 (s+1) 3 Q = T 1 T 1 (2) T 2 = 4(s+1)3 5(s+3) Robust Control 41
42 Nevanlinna-Pick Problem Problem: Let {a 1,...,a n } be a set of points in the open RHP and {b 1,...,b n } a set of distinct points in complex plane. Find a stable, proper, complex-rational function G satisfying: G 1 and G(a i )=b i, i =1,...,n Solvability: The NP problem is solvable iff the n n Pick matrix Q, whose ijth element is 1 b ib j a i + a j is positive semidefinite (Q 0). Note that Q is Hermitian (Q = Q where Q is the complex conjugate transpose of Q). Q 0 iff all its eigenvalues are 0. Mobius Function: A Mobius function has the form: M b (z) = z b 1 zb where b < 1 M b has a zero at z = b and a pole at z =1/b so M b is analytic in open unit disk.. M b maps the unit disk onto the unit disk and the unit circle onto the unit circle. The inverse map M 1 b Model Matching = z + b 1+zb = M b is a Mobius function too. 42
43 Nevanlinna-Pick Problem NP problem for n =1: Find a stable, proper G(s) such that G 1 and G(a 1 )=b 1 where b 1 1 and Re a 1 > 0. Case 1 b 1 =1: The unique solution is G(s) =b 1. Case 2 b 1 < 1: The set of all solutions is: {G : G(s) =M b1 [G 1 (s)a a1 (s)],g 1 CRH, G 1 1]} where the all-pass function A a (s) := s a s + a Example: For a 1 =2and b 1 =0.6 we have: G(s) = G 1(s) s 2 s G 1 (s) s 2 s+2 G 1 (s) =1results in G(s) = s 0.5 s +0.5 Remark 1: If G 1 is an all-pass function, so is G Remark 2: When a i are the complex-conjugate pairs, if G = G R + jg I is the solution of the NP problem then G R is also a solution to the NP problem. Model Matching 43
44 Nevanlinna-Pick Problem Consider the NP problem with n points: Case 1 b 1 =1: G(s) =b 1 is the unique solution (and hence b 1 = b 2 = = b n ). Case 2 b 1 < 1: Pose the NP problem with n 1 data points: {a 2,...a n } and {b 2,...,b n} where b i := M b 1 (b i )/A a1 (a i ) i =2,...,n Lemma: The set of all solutions to the NP problem is G(s) =M b1 [G 1 (s)a a1 (s)] where G 1 (s) ranges over the solutions to the NP problem. Example: Consider the NP problem with a = {1, 2} and b = {1/2, 1/3}. Solvability: The problem is solvable, because Q = 1 b b 1 b 2 2a 1 a 1 +a 2 1 b 2 b 1 1 b 2 2 a 2 +a 1 2a 2 Model Matching NP problem: a 2 =2, b 2 = NP problem: G(s) = = 3/8 5/18 eig(q) =[ ] Q 0 5/18 2/9 b 2 b 1 1 b 2 b 1 a 2 a 1 = 0.2 a 2 +a 1 1/3 = 0.6 G 1(s) = s 8 s 1 s+8 s s 8 s 1 2 s+8 s+1 = s2 3s +8 s 2 +3s +8 s 2 s s 2 s+2 = s 8 s +8 44
45 Model Matching Problem Find Q such that γ opt =min γ { T 1 T 2 Q γ} Define: G = 1 γ (T 1 T 2 Q) We find first G such that G 1 then we compute Q = T 1 γg stability of Q, T 1 γg should contain the RHP zeros of T 2 (i.e. z i ), that is: T 2 γg(z i )=T 1 (z i ) G(z i )= 1 γ T 1(z i ). However, to ensure the This is a NP problem and γ opt is the smallest γ for which the problem has a solution. That is, the associated Pick matrix is positive semidefinite. A γ 2 B 0 where : A ij = 1 z i + z j B ij = T 1(z i )T 1 (z j ) z i + z j Lemma: γ opt equals the square root of the largest eigenvalue of the matrix A 1/2 BA 1/2. Model Matching 45
46 Model Matching Problem Procedure: Given T 1 and T 2 find a stable Q to minimize T 1 T 2 Q (T1=tf(num,den)) Step 1: Determine z i the zeros of T 2 in Res >0. zz=zero(t2);z=zz(find(real(zz)>0)) Step 2: Form the matrices A and B: A ij = 1 z i + z j B ij = T 1(z i )T 1 (z j ) z i + z j Step 3: Compute γ opt as the square root of the largest eigenvalue of the matrix A 1/2 BA 1/2. gamma=sqrt(max(eig(inv(sqrtm(a))*b*inv(sqrtm(a))))) Step 4: Find G, the solution of the NP problem with data: z 1... z n γ 1 opt T 1(z 1 )... γ 1 opt T 1(z n ) Step 5: Set Model Matching Q = T 1 γ opt G T 2 Q=minreal((T1-gamma*G)/T2,0.01) 46
47 Model Matching Problem State-Space Procedure: Step 1: Factor T 2 as the product of an all-pass T 2ap and a minimum phase factor T 2mp Step 2: Define R := T 1 and factor R as R = R 1 + R 2 with R 1 strictly proper with all poles in T 2ap RHP and R 2 H and find a minimum realization of R 1 (s) = A B C 0 Step 3: Solve the Lyapunove equations: AL c + L c A = BB A L o + L o A = C C Step 4: Find the maximum eigenvalue λ 2 of L c L o and a corresponding eigenvector w. Step 5: Define: f(s) = A w g(s) = A λ 1 L o w C 0 B 0 Step 6: Then γ opt = λ and Q =(R λ f(s) g(s) )/T 2mp Model Matching 47
48 Design for Performance Objective: Find a proper C for which the feedback system is internally stable and W 1 S < 1 Lemma: If G is stable and strictly proper, then lim τ 0 G(1 J) =0where J(s) = P and P 1 stable: In this case the set of all stabilizing controller is: 1 (τs+1) k C = Q 1 PQ Q H and W 1 S = W 1 (1 PQ) Clearly, Q = P 1 is stable but not proper, so let s try Q = P 1 J to make it proper. Then W 1 S = W 1 (1 J) whose -norm is less than 1 for sufficiently small τ. P 1 stable: Do a coprime factorization of P = N/M, NX + MY =1 Set J =(τs+1) k with k = the relative degree of P Choose τ so small that W 1 MY(1 J) < 1 Set Q = YN 1 J and C =(X + MQ)/(Y NQ) Robust Control 48
49 P 1 Unstable (General Case) Assumptions: P has no poles or zeros on the imaginary axis, only distinct poles and zeros in the RHP and at least one zero in the RHP. W 1 is stable and strictly proper. Procedure: Step 1: Do a coprime factorization of P = N/M, NX + MY =1 Step 2: Find a stable improper Q im such that: W 1 S = W 1 M(Y NQ im ) < 1 It is a standard model matching problem that can be solved using the NP algorithm. Step 3: Set J = Step 4: Set Q = Q im J 1 with k = large enough that Q is proper and τ small enough that (τs+1) k Step 5: Set C =(X + MQ)/(Y NQ) W 1 M(Y NQ im J) < 1 Design for Performance 49
50 Design Example Flexible Beam: Consider the following simplified plant transfer function: P (s) = 6.47s s s(5s s s ) zeros Performance Specification: Settling time 8s and overshoot 10% poles ± 5.27j Assume that the ideal T (s) is a standard second-order system: ( ) ωn ζπ T id (s) = s 2 +2ζω n s + ωn 2 8 exp ζω n 1 ζ 2 =0.1 ζ =0.6 ω n =1 Then the ideal sensitivity function is S id (s) =1 T id (s) = We take the weighting function W 1 (s) to be S 1 id (s): W 1 (s) = s2 +1.2s +1 s(s +1.2) stable, strictly proper = W 1 (s) = s(s +1.2) s s +1 s s +1 (s )(s +1.2)(0.0001s +1) Design for Performance 50
51 Design Example Step 1: P (s) has a pole on the imaginary axis (s =0) so we perturb P to fix the problem (we add 10 6 to the denominator) Step 2: The model matching problem is to minimize: W 1 S = W 1 (1 PQ im ) P has only one RHP zero at 5.53, thus min W 1 (1 PQ im ) = W 1 (5.53) =1.02 and the specification is not achievable. Step 3: Let us scale W 1 as W 1 := W 1. Then the optimal Q im = W W 1 P 1 Step 4: Set J(s) = (τs+1) 3 and compute W 1(1 PQ im J) for decreasing values of τ τ norm Step 5: C = Design for Performance take τ =0.04 and set Q = Q im J Q 1 PQ = (W 1 0.9)J W 1 (1 J)+0.9J P 1 51
52 2-Norm Minimization Objective: Given P and W, find a proper stabilizing controller to minimize the 2-norm of a weighted closed-loop transfer function: e.g. min WPS 2 Define: The subspace of functions in L 2 that are analytic in the open RHP (all poles with Res 0) is the orthogonal complement of H 2 and is denoted by H 2. Every function F L 2 can be expressed as F = F st + F un where F st H 2,F un H 2 Lemma: If F H 2 and G H 2, then F + G 2 2 = F G 2 2 Problem: Obtain Q H to minimize WPS 2 = WNY WN 2 Q 2 Idea: Factor U := WN 2 = U ap U mp, then we have: WNY WN 2 Q 2 2 = WNY U ap U mp Q 2 2 = U 1 ap WNY U mpq 2 2 which leads to: Q im = Ump 1 1 (Uap WNY) st = (U 1 ap WNY) un +(U 1 ap WNY) st U mp Q 2 2 = (U 1 ap WNY) un (U 1 ap WNY) st U mp Q 2 2 To get a proper suboptimal Q, Q im should be rolled off at high frequency. and the minimum of the criterion: (U 1 ap WNY) un 2 Design for Performance 52
53 Optimal Robust Stability Objective: Given P ɛ =(1+ W 2 )P where ɛ, find the controller C that stabilizes every plant in P ɛ and maximizes the stability margin: γ inf := inf C W 2T ɛ sup =1/γ inf Procedure: Input P and W 2 Step 1: Do a coprime factorization of P = N/M, NX + MY =1 Step 2: Solve the model-matching problem: W 2 T = W 2 N(X + MQ) with T 1 = W 2 NX T 2 = W 2 NM and find Q im and ɛ sup =1/γ opt Step 3: Let ɛ<ɛ sup and set J(s) =(τs+1) k where k is large enough that Q im J is proper and τ small enough that: W 2 N(X + MQ im J) < 1 ɛ Step 4: Set Q = Q im J and C =(X + MQ)/(Y NQ) Robust Control 53
54 Robust Performance Problem Objective: Given P, W 1,W 2 find a proper controller C so that the feedback system for the nominal plant is internally stable and that: W 1 S + W 2 T < 1 This problem cannot be solved! Modified Problem: Consider the following inequality: W 1 S 2 + W 2 T 2 < 1/2 The robust performance problem with this inequality can be converted to a model matching problem (See Feedback Control Theory chapter 12.3) This inequality is a sufficient condition for the inequality in the exact problem. General framework: The inequality in the modified problem can be presented also as: W 1 S W 2 T =max ω σ max W 1S(jω) W 2 T (jω) < 1 2 Robust Control 54
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